ExtraSolvedProbs

ExtraSolvedProbs - Extra Solved Problems 7.11, 8.2,...

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Unformatted text preview: Extra Solved Problems 7.11, 8.2, 115,126,127,1312,1313, 15.13, 16.5, 192,193 Problem 7.11 An industry has a we sic stream consisting only oftlte organic chemical propanoic acid (informaily, propiouic acid}. Pmpanoic acid. CHsCHECOOH, undergoes chemistry similar to acetic acid. Its conjugate base, Cl'l3CH3COO”, is called propanoatc. A. When a 10" M pt'opaucic acid solution is prepared, the equilibrium pH is 3.97. Find the equilibrium constant for the dissociation ofpropzmoic acid to propanoate. B. The waste stream (consisting only ofpropancic acid and propanoate in water) is at pH 4.7. Using your answer from part A, find the total propanoic acid concentration {m [CI-IECI-iZCOOH} -i~{Ci-I,CH,CDO'] }. Purpose of problem Equilibrium calcuiation with a simple system Relevant sectiowts) of text Sections 7.3 and 7.4 Solution A: . The system is (assuming activities and concentrations are interchangeable): Unknowns: [H30], {PR}, [OI-1'], [9"], and [HF] whom: P‘ w propanoctc =Cli3Cii2C‘OO' and HP m propanoic acid = 01130450011 Constminis: Equiiibt'ia: KW= {I-FJIOI-I‘Pfi-IIO] m 10'“ K“: {I’fll'i’Ffi-IPJ Mass balance: Pt= [9'] + {HP} Charge balance: [13*] m [P‘} + {OI-F] Other: {1130} m 1 Simplified system: [H‘HOH‘] m K“, [1”]{1‘13551‘3’1 “15 PT? {13-} + 1511?] {if} m + {OH—I You know the system is acidic, so the charge baiance becomes: [I-l‘} *4 {P“]. From cq. 7.8: .{P‘} = KPflfI-l‘] -‘r K). So: [H43 “ EFT/([14+] “9 K) C“) Soiving fer 113K = {WETPT — [11+]) At pH 3.9.7, 31*] = 10-3“. Aiso, Pr == 10-3 M. Substituting: K = 1.29x10-5 (pK = 4.89) Without approximations: {HT + 1({1-1‘12 — (PTK + KWHH‘] - m“, = 0 (**_) SOiVing for K: K= - ({H”]3 + KWU'F] M [H’]2 — Pfiflé] - KW) = 1.29X10'5 B: Soiving {*3 for P1: Pr= [H‘KfH‘] + $35K With {14"} «a 10"” and K: 1.29x10-5:1>,= 5.1xm'5 M Wiflmut approximations, solve (* *) for PT: PT= ([1113 + RTE"): - 16,934+} — KKWHMIHHB = 5.1X10'5 M Problem 8.2 An engineer is designing a ohlorino contact chamber to disinfect poultry processing wastes. (Note: These wastrel: are often contaminated with the pathogenic bacteria Salmonella). She can design the system to use chlorine gas (equivalcm. to using HOCl) or liquid bleach (equivalent to using sodium hypochloritc, NaOCl) as the chlorine source. The total cltlonno concentration to be addocl is 7,] ingfL as Ci2 (= 10“ M chlorine). The species ilOCl is a much. strongor disinfectant than the species OCl‘. ’l‘i'it-:,1:U'C_I for HOCl is 1'54. A. ll“ chlorine chemistry controls the pH: which systom provides more disinfection? (Hint: This problem requires you to calculate equilibrium monocutmtions ofl-lOCl and OCl' in 10“ M I-lOCl and it)“ M NaOCl systems.) B. in a (sample-x waste stream such as this, is the assumption iltai “chlorine chemistry controls the pH" reasonable? Why or why not? Purpose of problem Graphical solutions of acids and salts Relevant section(s) of text Sections 8.4 and 8.5 Solution A: Tho pC-phl diagram can be sketched by the approach in Example 8.4, assuming ‘NaOCKsj dissolves completely. The [3Kw of l-lOCl is 154. Thus. the system point is at pC =2 total chlorine =- 4 and pH = 13K” «~= 7.54. The pC~pH diagrams for the HOCl and NaOCl systmcs are identical, except the NaOCi system also should have a line: representing [Naif]. Tho pC-pli diagram is sketched below, with [N5] not shown. The concentrations of HOCE and OC]‘ can be tound by determining the equilibrium p13 for each system. Use the proton condition, since NaOCi is a salt. The spooios list is: HBO, ii", OH‘, (3%“ for the NaOCl system only), OCl", and HOG]. For the HOG! system; starting materials: irl 0C1 and H20 proton condition: [H‘] [OH‘] + [OCH (same as the charge balance} oqu iiibrium pH: about 98 5.7 (where [If] m [CC]? and [OH‘] is small) For the NaOCI 3351mm Starting maicrlals: NaOCl a sz + OC-l“ and EEO proton condition: {8‘14- [HOCI] = [OH‘] equilibrium pH: about pH 8.8 (where [014"] v: [l-lOCl} and [W] is small) I! is clear from the pC-pl-l diagram that [HOCI] >7) [OCT] 21! pH 5.7 and [OCl‘] 7» [HOCi] at pH 8.8. Since HOCI is the stronger disinfcclam. the HOCI must he the stronger disinfecting sulutinn. B: Chlnriue mnlmls the chemistry here if the added chlorine species determine the final pH. Since only a small amount (10“ M) of a weaklyodisaociating acid (Kc, = 10”“) is added, it is unlikely that chlariue will control {he pH. Problem [1.5 An industry is discharging an acetic acid waste stream. (Recall that arctic acid is CHICOO‘H.) Their Nafianal Pollutant Discharge Elimination System (N'PDES) pulmjl rcquireg pH % 4 and total organic carbon (TOG) < 250 mgfas C. What is the maximum sirenth of acclic acid mste (as TOC in mgfl.) that can be discharged without vialafing either pennit requimmeni? (Recall that TOG, in mgfL, measures the number (11‘ milligrams ofcarbon per liter.) [’urpnsc ol‘problem Equilibrium calcuimiou with anmnoprotic acid Relevant section(s) of 1211 Section 1 1.3.6 Salution The Halal mass of 250 nag/L corresponds lo Acr= [HA9] + [Ac‘ ] = (0.25 g CELVHIE- g Cimnl 00 mo} C‘Jmo} total aocimcn =fi 1.04XE0'3 M. The simplifiud sysLem is as follows: Unlmmxm: {If}, [OI-1‘], [HAc}_._ and [Ac’] Equations: Ky; [mom K = [Ac-jgmllm m 10-“ ACT -= [HAG] + [Ac’] [91"]= [AF] +1031 At pli near 4, the charge balance becemes: {11"} m {Aer}. Recall that [Ac‘] :At‘pK/(X + [I-FJ), so: [tr] Maggi}: + [15mm {11*}! + [{[H‘] - mm o If alarm-1 .04XI0‘2 M, then UT] = 10‘3"“ or about pH 3.4. Thus, discharging rigfl at the TOC discharge limit will violnie the pH limit. How much acetic acid could be discharged 10 meal pH 4? At pH 4, ACT: [WEN + [[AI‘BIK m 6).le 10“ M. This is below the T00 discharge- limll. Nola 1l1al Acrm {6,0} XIO" moi total acclateiLXZ moi Cfmol total acetatak 12 g Cfmol C) = £44 mgfL C. Thus, abaut “l4 mg.’L TOC (6.0]X18"‘ M total acetate} could he disclialged and the pH will in: above 4. I’mhlem 12.6 When adding a strong base to an acidic wasie stream, the industrial wastewater treatmeht technician noted that the pH did not change much when a little more base was added. At this point, he had added 2 gaiions of l M NaOH to 750 gallons of waste. The pH was about 5.8. Estimam the p16, and concantraiiou of acid in the original waste (assuming it is monoprotic). Purpose of problem Relevant secfion{s} of text Section Snlu (ion The 9!“! does not change very much (La, the solution is weli buffered) atf= 0.5. This coma-spends to C3 -'= 16C and pH =2pKa. Here: CE =2 (E M NaOI-DCZ gallonsyflfi toia! gallons) = 2.66X10‘3 M. So: C »*~'u."al(2.66><10'3 M) a 5.3Xil)‘3 M. The 12K, is about equal to the pH = 5.8. Thus: C : 5.3><1(}‘3 and pK, = 5.8 a : i’roblcnl 12.7 An industry discharges a waste stream which consists of a singie monoprotic acid calicd HX. The titration curve for 1116 titration of the waste steam with N210?! is giver: below. Unfoflunalciy, the technician forgol to record tho PEX concentration in the waste stream, the 'NaOI-I concentration of the titmnt, or amount ofbasc added. (In the figure below, the tick marks are evenly spaced and the leftmost lick mark con'csponds to zero base addod). A. Estimate the 11K; of HX. B. What is the 10121] HX concentration (= {HX} + [X"]) in Lhe'wasic stream? C. How many liters of i M NaOH are required to neutralize 100,000 I. of the waste 10 pH 7? 13 12 {1 £0 a a / T / Base #fidedi?) Purpose of problem Relevant section(s} of text Section Solution A: The 13X“ can be cstnnated by the pH at Lhehaif—cquivaicncc poini. This is the first flat portion of the titration curve near pH 7. Thus. the pK, is about 7. ‘\ - x r Before any base is added, the pH of the waste is 22:01:14.5 fiv-intercept ofihe titration curve). Thus, C M '0fo (pKn about 7) givm a pH of 4.5. The charge balance for an HX soiufion is: ‘ [W] = [X’] + [03'] W {W} “C ’g/(EH'? + K") + {0H} Tim: C: (W3 - [OH'D'LUT] + KJ/Ka. Solving at pH 4-5 and qu m 7: C= 0.01 M. Sn the toad HX canccntration in the waste is about 6.01 At any point in the fila‘ation: c3, = [A‘] + {014"} - {m -= a,C+ [OH-1 — {W}. At pH 7: C, = OLEC = CKg’Cfiij +19.) = (0.01 MK] 0"}!(10'? '1' 10"?) = 0.005 M. To caicuiale the mining required: 196%.“, = VDCB. With QM = I M, V0 a 100,000 L, and (‘3 a 0.005 M, v = 500 L. 513 500 L of l M NaOI-I are required to neutralim the waste stream to pH 7. i it E i! J :i Problem 13.13 In an anaerobic digcsler, organics: in wastewater sludge are con varied into volatile acids {short- ch air: ()rganic acids), which are subsequently converted inio methane and carbon dioxide by a group of organisms called :rmifmgenx. in a typicai anaerobic digesier (pH 7, AH; m 3000 mgi'L a5 CaC03, volatile acids concentration “5 250 mgl’L}, how much of the alkalinity is due to the weak vulaliie acids? Assume that the volatile acids are acetic acid with 9K, 4.21. Purpose [If prairie-m Applications 0f alkalinin concepts Relevant seciion(s) of text Seclion 13.6 Soiution From Table 13.2, the zero [wait of protons for monoproflc acids with pH, «1 4.5 is HA. Thus, the alkalinity equation becomes: Aik = (or; + 2.3230,. + [03"] — [W] + [M] At pH '3‘ and with pKfl 4.7 andArw (0.25 giiyfffif) gimme) =4.17><10" M, [Ac'] w4.15><10‘3 M (Note: {Ad} z AT since pH 26> pfifa} Since acetate has a charge 0f~ l, 4.15%“)‘3 M-fl 4.15X19'3 cqiL or 207 mgfl. ofCaCOJ. Thus, aceiatc accounts for about 297:3(300 = 7% or the alkaiinity. Problem 13J5 ' Draw a titration curve like Figures 13.6-13.8 [0 showthe finctions ofacidity. Purpasc of probiem Understanding of acidity Relevant secfion(s) of text Section 13.4 S elation The titration curve- is shown below for a very acidic water (to illustrate each acidity fraction) [IIth equlm 1St equiv. 2”” equiv. pint pint '11me with acid titrate with base } 3-1 :1 Problem 15.!3 A community uses a groundwater supply for drinking water. The water is fairiy hard, wi :11 a hardness of 300 mgfL as C3C03. To minimize soap scum, ihc residents wish io add a dcie-rgcni containing nitriloiriacetatc (13‘) to iiieir washing machines. The detergent is 3% U" by weight i'fthc average washing machine has a capacity of 40 L, how much demrgem should be added to reduce the hardness to 25 mgfLas CaCOJ? Assume the hardness is from calcium. The molecular weight of If” is 138 g/‘moie. Furpose of problem Compicxation with an organic ligand Relevant séctioMS) of text. Section 2.5.4.4 Solution The porlioent equilibria are (from Appendix D): 1-131. = HIL‘ 4* Ir KI = 10" 1" a}; = 1-113" + W .ic2 E1045M HI?" = 1'}- Hi“ K3: 104‘!” Cay +L3” a CaL‘ K4 "mi 10“"3 CaEWZLJ' meal;- K3='JO"-“ You do not know 111:: pH. Since this is treated drinking walcr, you can assume that the 511i is near neutral. Thus, ihc only importani acid-base. form of L is L3”. Also, CaOH‘ can be ignored (see text). The mass balanccs are: Cerium [Ca1+] {Cari + [Calf] w [Caziiu + Km] + Kim?) (*i LT: {Lil + [Cat] + mam -~= n?" 1(1 + K1921 + KgiL’JiCaRD no We want to find I.3r so that {C25} is fixed ai the dosired hardness. The desired hardness is: desired hardness = (25 mgfL as CaCOEMSGQOO cq!L per mgfL as CaCOJ) = SXiO‘ 4 (2qu So: [032‘] = (SHIV equ}1(2 equot) = 2.5XIG“ M You know Ca{+ ID, from the initial Madness: initial hardness w (300 mgi’L as CaCOQKSOflGO (2qu per mgiL as C3603) m 6X] 9* cqu- So: CM +11), = initiai {Cak} == (6X if)“ eqiL)f(2 eqi’mol} = 3% 113“1 M x ‘x x x i ‘ \ The spa-mach is to solve (*) for [1.7"] (with Cari-MT: 3x if?" M and [Cay] =2.5><10“‘ M). After [1.3"] is found, LT can be caiculaied fi‘om (**}. Soiving 0‘} for {L’}: 10"3‘{L3']2 + 10am] — 11 = e or [If] = 5.50540“ M. Thus: L,= {3.316 +K4[Ca2*] + K5[L3"][Ca’*}) = 2.76x10‘3 M L,» required : (2.76X10'3 M)(188 gimoie) = 0.518 gfI. m (0.518 gIL)(40 L) a 203 g M40 1. detergent required = (20.7 $10.03 = 691 g detergent for 40 L Grabth 1.5 pounds ofdeiergenl per load - Note: 1., is large enough that activities and concentrations may not be fitrcrchangeable. Problem 16.5 A drinking water treatment plant is using ozone for disinfection, Plant personnel want to use ozone also to oxidize reduced manganese, ani to MnOJs). i A. Find :he required ozone: dose {in g ()3 per 5; Mn" oxidized} if ozone is reduced to oxygen. 13. Plant personnel added ozone to their water and noticed a pink water condition. What was I the minimum. ratio 0103 lo Mn” that they were using? Purpose of problem Detour: i11ng doses from overall redox reactions Relevant secfion(s) of text Secu‘on 16.3.4 Soiul‘iou A: 03(aq} + 5119* -‘ M13015!) + 0203(1) Oxidation halfrcamion: Mt:2+ + 40H — MoOfi} + 21-120 + 26' Reducliun hair reaction: 03mg) + H20 + Ze' a 02am) + 20H” Overall: 05 4: Mn +~fr 20H — Mnogs} + 02 4» H20 ' The dose is I mole 0311 mole May or 48 g 0364.9 g Mn“ 23 0.87 g 03 per 1.; Mn!“ oxidized Pink moans permanganate formation (MIKE): 03(aq) + Mn" - Mm); + 03(aq) Oxidation haifrcfidiou: Mn” + 80H - MnO; + 4H4) + Se“ Reduction hzll'reaction: Ogaq) l-IEO + 23‘ -~ 043(1) + ZOH' Overall: 505(aq) + 2M?» + 60H —‘ 2M n04" + 502(aq) + 31430 The minimum dose is 2.5 mole 03:1 mole Mn” or (2.5 mol)(48 glmol 09554.9 g Mitzi or 2.2 g ()_1 per g Mn“ i’rohlem 19.2 A municipai wastewaier at pH 7.5 contains 6 mgr’L as 'P of totai oribophosphau: (i-e.,P(+V). It is desired to precipitate the phosphate to achieve a residual orthopliospliate concentration of 0.5 ingr'L as P using ferrous iron. (For Parts A through C, ignore coniplexarion/precipitafion reactions between ferrous iron and ligands other than P043”. Assume that the Fe'fi-PID4 soiid phase has the same composition as vivianite, but is 107 times more soiuble than Vivienne.) A. How much feirous iron must be added to satisfy the sicichiometty of die precipitation reaction? (See Appendix D for die appropriate equiiibria.) B. How much ferrous iron must be added to maintain equilibrium between the residual phosphate and the solid phase? ' C. The required dose is the sum of the ferrous iron doses calculated in Pans A and B. What is the required dose? How much total ordiophosphaie would have been removed if only The dose based on stoiehiomelry (Le, your answer to Part A} had been added? D. Under the water quaiin conditions stated in the problem, what soiid controls the soiubility: Fe{Ol-l}2(s) or Fe3(PO_I)2(HzO)a? Purpose of problem Controlling solids and solid—liquid equilibrium caicuiations Relevant section(s) of text Sections 19.3 and 19.4.2 Soluu'on A: The solid 1' armed is FegPOJLUiIOh. Thus, each mole ofphosphaie removed requires 3:? = L5 moles of iron ‘ So: Fe“ dose for precipitation = (1.5)(10tal orthophosphaie removed) ' m {1.5% ~ 0.5 mgi'LMBlfiGO mg’mol} 3 2.58349" M Note: it is ok to use totai orthophosphare here, because the acidfbase system will re— equiiibrate as P0; is removed. Thus, ail orthophosphale species wili be removed. B: From Appendix D: Felipoqm) m 3Fez*+ 2pm“ flaw-107K w mam savifianiae Thus: [Fc2*]3[P0‘-’>"32 =- [(50 w 1040.25 So, to maintain the cquiiibrium with the solid: [Ff] m {KmEIPOf'Fim ’5 {Kg/(dag)? ["3 m3 = 8.39“ 0"“ at pH 7.5 Pr” residuai orthophosphate = (0.5 mgj’L as P):’(31 mgfmmol) == 1.61X10‘5 M So: [Fey] m {]0‘3"“1’fl839x10'6){I.6ix10" 5 Mm“? = 2.12x10‘4 M If other compicxes are ignored, then the ferrous iron dose to maintain the equilibrium with [he sniid is 2.12x1'0" M. A giancc at ihe smhility consiants for Fey—OH" samplexes reveals that Fish-0H compiexes can be ignomd at pH 15. The requin dose is: dose from stoichiometry + dose to maintain equilibrium = 2.58¥10"“M + 2. [2x 10“ M = 4.70)( 10“ M What if only the stoichiomclric dose (12.58310"1 M; was added? dose added m dose finm stoichicmcny + dose to maintain eqfiiiibn'um = 1-503-110 ‘ P r) + {K201 (“310 232} :3 Solving forPT with dose added = 2.58“ 0‘4 M and PH, = initial orthophosphate 7* 6 mgi'L as P = 1.94x’10“ M: I"r H 7.34MB“ M = 2.3 mgIL as P. This is only 62% removal, not 92% rcmovai as required. The pertinent equiiibria are: FGSCPOQAS) = 3Fe”+2?043' K50 = 107K slink-innit: Fe{OH)J(s) = Fe“ + 2014' 1&1,mu = 30"“43 = i 0- .3036 At pH 7.5: [933} in equiiibrium with Fe(OH}2{s) m K'w‘thOI‘I' 32 = 104"” M = 331% 16'2 M At pH 7.5 and residua} PT“: LGixlfi" M: [R335] in equilibrium with 13630304)2(S)fi 2'.12><1{}‘4 M (from Part B} ' Thus, Fe3(I’04)2(s) controls the ferrous imn scalability, since it gives the smaller equilibrium {FURL ' Problem 19.3 A student is studying a hazardous waste stream containing cadmium and she needs to know the. solubility product fnr CdCOJs). She adds some Cdcoats) to water and measures the pH as pH 7.1. The solid does not dissolve completely. Assuming that the suspensitm is in equilibrium: with the atmosphere (PC02 = 10‘” atm), What is the if,“ for Cdcogts)? Compare your answer with the K,“ for CdCOfistviw) (log zn= - 12.1 from Appendix D). Hint: Perform a charge baiance and think about the alkatinity equation in an open sy$tem. I Purpose of prnblem Soiid~liquid equilibrium calculations I Relevant section(s) 13f text Section 19.3 Solution The pertinent equiiibrium is: CdCO-sfis) m Cdz" + (3033' K," Peiforming a chm-gt: balance: 2MP] + {14*} = [Hang] + 2[c03=*] + [OH' } Why only these species? _ From the equiiibrium comtants in Appendix D, it is apparent that C‘d(0}-I),f“-r are small relative to Cd?“ at pH 7.1 - Aiso CdHC‘Of is important relative to Cd“ only if [HCOg] > 1(3"m M or so (from th equilibrium: Cd” + HCOE,‘ fl CdHCOf ; K = 10””). However, [HGOa'] == an 1C," a alKHszma = KIK, wqu‘] << 1043-9 M at pH 7.1. The charge balance becomes: 2tc‘d21 = (a1 + zettwmmn + {OH} — Em Bm: [Odfll =Kmflcozz-1 = Ksflazcr) g thamufiazKHPL-oz) So: 2Kmote/(et2K; m2} = (tit.1 +2a2}(KHPCD1/‘au)+ [OH'] - [if] Solving for Km at pH 7. i , with P901 = 10‘3-5 atm, and KHr-w 10"'51noLiI.-<ztnt: K50 a 1.26% 10~12 or lt)g[i',.J = w l 1.9(cinse to —12.l} It is remarkable that a Kfl value can be estimated by a single pH measurement! ...
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This note was uploaded on 02/02/2012 for the course CEE 255B taught by Professor Michaelstenstrom during the Fall '11 term at UCLA.

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ExtraSolvedProbs - Extra Solved Problems 7.11, 8.2,...

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